Solution: Without loss of generality $x$ is a natural number $n$. We prove the claim by induction on $n$. If $n = 0$ then the claim holds because the only function $0 \to 0$ is the empty function, which is a bijection. Now suppose that every injection $n \to n$ is a surjection, and let $f: n+1 \to n+1$ be an injection. Consider the restriction $f \restriction n$, which is also an injection.
Case 1: If $\text{ran}(f \restriction n) \subset n$ then by the induction hypothesis $\text{ran}(f \restriction n) = n$. Therefore in this case we must have $f(n) = n$, so $\text{ran}(f) = n+1$ as desired.
Case 2: If $\text{ran}(f \restriction n) \not\subset n$ then $n \in \text{ran} (f \restriction n)$, say $f(k) = n$ where $k \lt n$. Then the function $g : n \to n$ defined by $$ g(i) = \begin{cases} f(i) & \text{if } i \lt k\\ f(i+1) & \text{if } k \le i \lt n \end{cases} $$ is an injection, so by the induction hypothesis it must be a surjection. This means that $f \restriction ((n+1) \setminus \{k\})$ is a surjection onto $n$, and because $f(k) = n$ we may conclude that $f$ is a surjection onto $n+1$.
Solution: Fix $\alpha$ and define a function $f : (\{0\} \times \alpha) \cup (\{1\} \times \text{Ord}) \to \text{Ord}$ by
The restriction $f \restriction (\{0\} \times \alpha) \cup (\{1\} \times \beta)$ is an isomorphism of $((\{0\} \times \alpha) \cup (\{1\} \times \beta); \lt_\text{sum})$ with $(\alpha + \beta; \in)$.
We show this by induction on $\beta$. The case $\beta = 0$ is clear.
Suppose the claim holds for $\beta$. Then the claim holds for $\beta+1$ because the "new element" $(1,\beta)$ in the domain maps to the "new element" $\alpha+\beta$ in the domain, and moreover both new elements are strictly maximal in their respective orderings: for every $(i,\gamma) \in (\{0\} \times \alpha) \cup (\{1\} \times (\beta +1))$ we have $(i,\gamma) \lt_\text{sum} (1,\beta) \iff (i,\gamma) \ne (1,\beta)$ and we have $(1,\beta) \not \lt_\text{sum}(i,\gamma)$, and for every $\xi \in \alpha + \beta + 1$ we have $\xi \in \alpha + \beta \iff \xi \ne \alpha + \beta$ and we have $\alpha + \beta \notin \xi$.
For the limit step, suppose the claim holds for all $\beta$ less than some limit ordinal $\lambda$. Then the range of $f \restriction (\{0\} \times \alpha) \cup (\{1\} \times \lambda)$ is the union of the ranges of $f \restriction (\{0\} \times \alpha) \cup (\{1\} \times \beta)$ for $\beta \lt \lambda$, which is $\bigcup_{\beta \lt \lambda} (\alpha + \beta)$ by the induction hypothesis, and this is $\alpha + \lambda$ because addition is continuous in the second coordinate. Now if the function $f \restriction (\{0\} \times \alpha) \cup (\{1\} \times \lambda)$ were not an isomorphism of $((\{0\} \times \alpha) \cup (\{1\} \times \lambda); \lt_\text{sum})$ with $(\alpha + \lambda; \in)$ then there would be some "problematic" pair of elements on which the function did not preserve the order. These two problematic elements would have already shown up at some stage $\beta \lt \lambda$, contradicting the induction hypothesis. It follows that $f \restriction (\{0\} \times \alpha) \cup (\{1\} \times \lambda)$ must be the desired isomorphism after all.
Solution Fix $\beta$ and define a function $f : \text{Ord} \times \beta \to \text{Ord}$ by
The restriction $f \restriction (\alpha \times \beta)$ is an isomorphism of $(\alpha \times \beta; \lt_\text{lex})$ with $(\beta \cdot \alpha; \in)$.
We show this by induction on $\alpha$. The case $\alpha = 0$ is clear.
Suppose the claim holds for $\alpha$. Then the claim holds for $\alpha+1$ because the function $f \restriction ((\alpha+1) \times \beta)$ is a composition of two isomorphisms: $$((\alpha+1) \times \beta; \lt_\text{lex}) \xrightarrow{g} ((\{0\} \times \beta\cdot \alpha) \cup (\{1\} \times \beta); \lt_\text{lex}) \xrightarrow{h} (\beta \cdot \alpha + \beta; \in) = (\beta \cdot (\alpha + 1); \in), $$ defined as follows. We define $g$ as the function that sends pairs of type $(\gamma,\delta) \in \alpha \times \beta$ to pairs $(0,f(\gamma,\delta))$ and sends pairs of type $(\alpha, \delta) \in \{\alpha\} \times \beta$ to pairs $(1,\delta)$. It is not hard to check that this function $g$ is an isomorphism using the fact that every pair of the first type is below every pair of the second type, and using the induction hypothesis to show that $g$ does the right thing with pairs of the first type. We define $h$ as the isomorphism from part (a) applied to the sum of ordinals $\beta \cdot \alpha$ and $\beta$.
The limit step is proved similarly to the limit step of part (a).
Solution: Let $\alpha = \text{otp}(A;\in)$. By the definition of order type, there is a (strictly) increasing bijection $f: \alpha \to A$. We may consider $f$ as an increasing injection $\alpha \to \text{Ord}$. Any such function must satisfy $f(\gamma) \ge \gamma$ for every $\gamma$ in its domain; this is easy to prove by transfinite induction on $\alpha$. (Alternatively, note that if $f(\gamma) \lt \gamma$ then $f(f(\gamma)) \lt f(\gamma)$ and so on, and we get an infinite decreasing sequence whose range is therefore a set of ordinals with no least element, contradicting well-ordering.) If $\beta \lt \alpha$ then the function value $f(\beta)$ is defined and we must have $f(\beta) \ge \beta$, which contradicts $f(\beta) \in A \subset \beta$. Therefore we must have $\alpha \le \beta$ instead.